Question

the mean percentage of a population of people eating out at least once a week is 57% with a standard deviation of 3.5 %. assume that a sample size of 40 people was surveyed from the population an infinite number of times. 95% of the sample mean occurs between ___ and ___?

Answer 1

The 95% confidence interval for the population mean based on a sample of 40 people is between 55.913% and 58.087%.

Step-by-step explanation:

The question involves the application of the Central Limit Theorem and concepts from statistics related to constructing confidence intervals for the population mean based on a sample. Given that the population mean is 57% and the standard deviation is 3.5%, and assuming we survey a sample size of 40 people, to find the 95% confidence interval for the sample mean we would use the standard error of the mean which is the standard deviation divided by the square root of the sample size. For 95% confidence, we typically use a z-score of approximately 1.96 (corresponding to the 95% confidence level).

The margin of error (ME) is calculated as:

ME = z * (sigma/sqrt(n))

Thus, the confidence interval is given by:

(mean - ME, mean + ME)

Plugging in the numbers:

ME = 1.96 * (3.5/sqrt(40)) = 1.087

So, the confidence interval is (55.913%, 58.087%).

Answer 2

Given:
Population proportion,
`\mu _(p)` = 57% = 0.57
Population standard deviation, σ = 3.5% = 0.035
Sample size, n = 40
Confidence level = 95%

The standard error is

`SE_(p) = \sqrt{ (p(1-p))/(n) } = \sqrt{ (0.57*0.43)/(40) } =0.0783`

The confidence interval is

`\hat{p} \pm z^(*)SE_(p)`
where

`\hat{p}` = sample proportion
z* = 1.96 at the 95% confidence lvvel

The sample proportion lies in the interval
(0.57-1.96*0.0783, 0.57+1.96*0.0783) = (0.4165, 0.7235)

Answer: Between 0.417 and 72.4), or between (41% and 72%)